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EMPIRICAL FORMULA
The simplest whole number ratio of elements in a compound is called the EMPIRICAL FORMULA. In contrast,
the molecular formula of a compound is the actual number of atoms of each element in the compound.
It can be calculated from the mole ratio or from percent composition data. If given percent data, consider it masses
out of 100 grams.
Mole Ratio Example:
What is the empirical formula for a compound if a 2.50 g sample contains 0.900 g of calcium and 1.60 g of chlorine?
Step One: Determine the number of moles of Ca and Cl. (You must first determine the atomic mass of each.)
Ca
0.900g Ca
1 mole Ca = 0.0225 mol Ca
40.08g Ca
Cl
1.60g Cl
1 mole Cl = 0.0451 mol Cl
35.45g Cl
Step Two: To obtain the smallest ratio, divide both numbers of moles by the smaller number of moles (0.0225 mol).
Ca
0.0225 mol = 1.00
0.0225 mol
Cl 0.0451 mol = 2.00
0.0225 mol
Step Three: Look at the ratio of the two numbers. Round off to whole numbers (there are exceptions to this).
Ca = 1.00 becomes Ca
Cl = 2.00 becomes Cl
1
2
The empirical formula is CaCl2
What do you do if you do not get nice whole numbers in your ratios?
• If the ratio is: X.1 or X.9, round up or down
• If the ratio is X.2 to X.8, determine what to multiply it by to get a whole number.
• Examples:
X.3 x 3 = X.9 which is rounded
X.5 x 2 = X.0 which is a whole number
X.4 x 5 = X.0 which is a whole number
MOLECULAR FORMULA
If you know the empirical formula of a substance and its molecular mass, you can determine its molecular formula.
1. Calculate the empirical mass from the empirical formula
2. Divide the molar mass by the empirical mass. Apply this number to all atoms in the empirical formula.
HYDRATES
-
A crystalline compound in which its ions are attached to one or more water molecules.
Hydrates are solids with water molecules trapped in them.
The formula is written as:
(the number of waters)
(the formula)
XY  XH2O
CuSO4  5H2O
copper (II) sulfate pentahydrate
DETERMINING THE FORMULA
It is a lot like calculating the empirical formula only you are not using single elements. You are using water and the
compound. Follow these steps:
1. Find the individual masses:
Original mass – anhydrous mass
=
water mass
Example: 5.00g
–
4.26g = 0.74g
2.
Find the empirical formula
4.26g BaCl2
BaCl2
1 mol BaCl2 =
208.23g BaCl2
0.0205mol BaCl2
0.0205mol = 1.00
0.0205mol
H2O
0.74g H2O
1 mol H2O
18.02g H2O
0.041mol = 2.0
0.0250mol
= 0.041mol H2O
BaCl2  2H2O
STOICHIOMETRY
This is the part of chemistry that deals with the amount of substances involved in chemical reactions (reactants and
products). Costs for consumer products are kept low because stoichiometric calculations allow scientists to increase
efficiency. Chemical reactions stop when one of the reactants is used up. Stoichiometry is the study of quantitative
relationships between amounts of reactants used and the products formed by a chemical reactions.
All stoichiometric calculations must start with a balanced equation. You can convert mole-mole, mole-mass, mass-mole,
and mass-mass.
Steps to follow:
1. Write down what you know and what you need to find out.
2. Write the balanced chemical equation.
3. Determine the number of moles of the given substance. You may have to convert it from grams to moles.
4. Determine the mole ratio between what is known and what is unknown.
5. Convert the unknown moles to grams if needed.
MOLE – MOLE CONVERSIONS
EXAMPLE:
How many moles of H2 are produced when 0.0400mol of Na is reacted with water ?
MOLE – MASS (or MASS– MOLE) CONVERSIONS
Practice:
Magnesium carbonate decomposes into magnesium oxide and carbon dioxide. If you are given 25.0g of
magnesium carbonate, how many moles of carbon dioxide gas are formed?
MASS – MASS CONVERSIONS
Example:
How much silver chloride, AgCl is produced from 17.0 g silver nitrate, AgNO3 with an excess of sodium
chloride, NaCl?
LIMITING REACTANTS
Rarely are reactants in a chemical reaction present in the EXACT ratios specified by the balanced equation. Usually
one or more are in excess while one is limited. The substance that is completely consumed (used up) in a reaction is the
limiting reactant. The reaction cannot continue without more of it. Excess reactants are the reactants that are left over, in
excess.
Treat these problems as mass to mass problems and then compare the answers to determine the limiting reactant.
The limiting reactant will be the one that converted to the lesser amount.
Example: How many grams of CO2 are formed if 10.0 grams of C are burned in 20.0 g of O 2.
PERCENT YIELD
We can calculate theoretical amounts of chemicals produced in a chemical reaction by doing limiting reactant
problems. In reality, the actual experimental amounts yielded in the laboratory are much less due to human error or other
factors. Chemists want to compare the theoretical yield to the actual yield to see how efficient they are.
Percentage Yield
=
actual amt of product x 100
theoretical amt of product
Theoretical amt – maximum amount of product that can be produced from a given amount of reactant.
Actual amt – amount of product actually produced when the chemical reaction is carried out.
Example:
In the reaction above, the actual yield of CO2 is 21.7g. What is the percent yield?
1.
The empirical formula of styrene is CH; the molar mass is 104.14g/mol. What number of H atoms are present in a
2.00g sample of styrene?
2.
Methane (CH4) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon
disulfide and hydrogen sulfide as the only products.
a. Write the balanced chemical equation for the reaction of methane and sulfur.
b. Calculate the theoretical yield of carbon disulfide when 1.20 x 10 2g methane is reacted with an equal mass
of sulfur.
3.
An ore sample contains Fe2O3 plus other impurities. A 752g sample of impure iron ore is heated with excess
carbon, producing 453g pure iron by the following reaction: Fe 2O3(s) + 3C(s)  2Fe(s) + 3CO(g). What is the mass
of Fe2O3 in the impure iron ore sample? Assume the Fe2O3 is the only source of iron and that the reaction is 100%
efficient.